The area (in sq. units) bounded by the parabola

Solution:

$\because \quad$ Curve is given as :

$y=x^{2}-1$

$\Rightarrow \quad \frac{d y}{d x}=2 x$

$\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(2,3)}=4$

equation of tangent at $(2,3)$

$(y-3)=4(x-2)$

$\Rightarrow y=4 x-5$

but $x=0$

$\Rightarrow y=-5$

Here the curve cuts $\mathrm{Y}$-axis

$\therefore$ required area

$=\frac{1}{4} \int_{-5}^{3}(y+5) d y-\int_{-1}^{3} \sqrt{y+1} d y$

$=\frac{1}{4}\left[\frac{y^{2}}{2}+5 y\right]_{-5}^{3} \frac{-2}{3}\left[(y+1)^{3 / 2}\right]_{-1}^{3}$

$=\frac{1}{4}\left[\frac{9}{2}+15-\frac{25}{2}+25\right]-\frac{2}{3}\left[4^{3 / 2}-0\right]$

$=\frac{32}{4}-\frac{16}{3}=\frac{8}{3}$ sq-units.